There are two vectors of equal magnitudes. When these vectors are added, then magnitude of the resultant is also equal to the magnitude of each of the two given vectors. Angle between the vectors is

Options

A. `120^(@)`
B. `60^(@)`
C. `30^(@)`
D. `150^(@)`

Explanation

To solve the problem, we need to find the angle between two vectors of equal magnitude when their resultant is also equal to the magnitude of each vector. Let's denote the magnitude of each vector as \( A \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
Let the two vectors be \( \vec{A} \) and \( \vec{B} \) such that \( |\vec{A}| = |\vec{B}| = A \).

2. **Resultant Vector Magnitude**:
According to the problem, the magnitude of the resultant vector \( \vec{R} \) is equal to the magnitude of each of the two vectors. Therefore, \( |\vec{R}| = A \).

3. **Using the Formula for Resultant**:
The magnitude of the resultant of two vectors can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
Substituting the magnitudes:
\[ A = \sqrt{A^2 + A^2 + 2 A A \cos \theta} \]

4. **Simplifying the Equation**:
This simplifies to:
\[ A = \sqrt{2A^2 + 2A^2 \cos \theta} \]
Squaring both sides gives:
\[ A^2 = 2A^2 + 2A^2 \cos \theta \]

5. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ A^2 - 2A^2 = 2A^2 \cos \theta \]
\[ -A^2 = 2A^2 \cos \theta \]
Dividing both sides by \( A^2 \) (assuming \( A \neq 0 \)):
\[ -1 = 2 \cos \theta \]

6. **Finding Cosine Value**:
Therefore, we have:
\[ \cos \theta = -\frac{1}{2} \]

7. **Determining the Angle**:
The angle \( \theta \) that satisfies \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \quad \text{(in the second quadrant)} \]

### Final Answer:
The angle between the two vectors is \( 120^\circ \).